On Schur Complement in k-kernel Symmetric Matrices

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1 Int. Journal of Math. Analysis, Vol. 4, 2010, no. 7, On Schur Complement in k-kernel Symmetric Matrices A. R. Meenakshi and D. Jaya Shree 1 Department of Mathematics Karpagam college of Engineering Coimbatore , India Abstract In this paper we give the Necessary and sufficient conditions for a Schur complement in a k-kernel Symmetric matrix to be k-kernel symmetric. Mathematics Subject Classification: 15A57, 15A15, 03E72 Keywords: Schur Complement, Kernel Symmetric,k-Kernel Symmetric 1 Introduction Let F mn denotes the set of all m n matrices over the fuzzy algebra F. In short F nn is denoted as F n. A development of the theory of fuzzy matrices were made by Kim and Roush3 analogous to that of Boolean matrices.throughout we deal with fuzzy matrices that is, matrices over a fuzzy algebra F with support 0,1 under max-min operations.for a,b F, a+b = max{a,b} and a.b = min{a,b}.for A F n, let A T,R(A) and N(A) denote the transpose, Rowspace and Nullspace of A respectively.we denote a solution X of the equation AXA = A by A.For a complex matrix M partitioned in the form M = a Schur complement of A in M denoted by M/A is defined as D CA B. This is called generalized Schur complement for complex matrices 1,7. A matrix A C n n is said to be k-ep if it satisfies the condition 1 shree jai6@yahoo.co.in

2 332 A. R. Meenakshi and D. Jaya Shree Ax=0 KA Kx = 0 or equivalently N(A) =N(A K). For further properties of k-ep matrices one may refer 6. For x =(x 1,x 2,...x n ) F 1 n and let κ(x) =(x k(1),x k(2),...x k(n) ). Let k-be a fixed product of disjoint transpositions in S n =1, 2,..., n and K be the associated permutation matrix.in this paper we discuss when a Schur complement in k-kernel symmetric fuzzy matrix will be k-kernel symmetric which includes the results found in 6 as a particular case and analogous to that of the results on complex matrices found in 7.The paper is organized as follows. In section 2, we give some basic definitions and results. In section 3 we discuss Schur complement in k-kernel Symmetric fuzzy matrices. 2.Preliminaries We shall recall the definition of k-kernel symmetric matrices studied in our earlier work5. Definition 2.1 A matrix A F n is said to be k-kernel symmetric if N(A) =N(KA T K). In the sequel, we shall make use of the following results. Definition 2.24,P.119 For A F n is Kernel symmetric if N(A) =N(A T ). where N(A) = {x/xa = 0 and x F 1n }. Lemma 2.34,P.125 For A, B F n and P be a Permutation matrix N(A) =N(B) N(PAP T )= N(PBP T ). 3. Schur Complement in k-kernel symmetric Matrices Throughout we are concerned with a block fuzzy matrix of the form M = (3.1) with respect to this partitioning a Schur complement of A in M is a fuzzy matrix of the form M/A = D CA B.where A and D are square matrices.here M/A is a fuzzy matrix if and only if D CA B, that is, D = D + CA B. A partitioned matrix M of the form (3.1) is k-kernel symmetric then it is not true in general that a Schur complement of A in M, M/A is k-kernel symmetric. Here the necessary and sufficient conditions for M/A to be k-kernel

3 On Schur complement 333 symmetric are obtained.throughout this section let k 1 and k 2 be the product of disjoint transpositions in S 2n defined as follows: For x = (x 1,x 2,..x n,x n+1,...x 2n ) k 1 (x) ={κ(x),x n+1,...x 2n } k 2 (x) ={x 1,x 2,..x n,κ(x)}, x n+1 x k(1), x n+2 x k(2),..., x 2n x k(n) If k = k 1 k 2 then k(x) =(k(x),k(x)) The permutation matrices associated with k 1,k 2 and k are K 0 In 0 K 0 K 1 = K 2 = and K = respectively. 0 I n 0 K 0 K In this section, k-kernel symmetric matrices of a block matrix is discussed. Theorem 3.2 Let M be a matrix of the form (3.1) with N(A) N(B) and N(M/A) N(C), then the following are equivalent. (1) M is k - Kernel symmetric matrix with k = k 1 k 2. (2) A is k -Kernel symmetric, M/A is k -Kernel symmetric, N(A T ) N(C T ) and N((M/A) T ) N(B T ) (3) Both the matrices and are k - Kernel symmetric. C M/ M/A Proof (1) (2): To prove A is k is k-kernel symmetric,m/a is k-kernel symmetric. Let x 1 N(A) and x 2 N(M/A).Hence x 1 A =0 and x 2 (M/A) = 0 (3.3) Define x = x 1 x 2 we claim that xm = x 1 x 2 =0 Since N(M/A) N(C), x 2 (M/A) =0 x 2 C =0

4 334 A. R. Meenakshi and D. Jaya Shree N(A) N(B),x 1 A =0 x 1 B =0 Hence, x 1 A + x 2 C = 0 and x 1 B + x 2 D =0. Therefore xm = 0 that is x N(M). Since M is k - Kernel symmetric, N(M) =N(KM T K) Therefore, xkm T K =0 (x 1 x 2 ) K 0 0 K A T C T B T D T K 0 0 K =0 x 1 KA T K + x 2 KB T K =0 x 1 KA T K = 0 and x 2 KB T K =0 and x 1 KC T K + x 2 KD T K =0 x 1 KC T K = 0 and x 2 KD T K =0 Hence x 1 N(KA T K),x 2 N(KB T K) and x 2 N(KD T K) Since x 1 N(A) and x 2 N(M/A) it follows that N(A) N(KA T K), N(M/A) N(KB T K) and N(M/A) N(KD T K) implies N(M/A) N(K(M/A) T K) Similarly it can be shown that N(KA T K) N(A) Thus A is k-kernel symmetric. Since, x 1 N(KC T K) and A is k-kernel symmetric N(A) =N(KA T K) N(KC T K) By using Lemma(2.3) N(A T ) N(C T ). By Definition M/A = D CA B implies N(M/A) N(K(M/A) T K) Similarly it can be shown that N(K(M/A) T K) N(M/A). Therefore N(M/A) =N(K(M/A) T K). Hence M/A is k-kernel Symmetric.

5 On Schur complement 335 Since,N(M/A) N(KB T K) and M/A is k-kernel Symmetric, N(K(M/A) T K) N(KB T K). Therefore by using Lemma(2.3) N((M/A) T ) N(B T ). Thus (1) (2) holds. (2) (3): M 1 = and M 1 = are k-kernel symmetric. C M/A 0 M/A Let x N(M 1 ). Partition x in conformity with that of M 1 as x = x 1 x 2 then, (x 1 x 2 ) =0 x 1 A =0,x 2 C =0;x 2 C = 0 and x 2 D =0 x 2 (M/A) = 0. Since A and M/A are k - Kernel symmetric, x 1 N(A) =N(KA T K) x 1 KA T K =0 x 2 N(M/A) =N(K(M/A) T K)) x 2 (K(M/A) T K)=0. Since, N(A T ) N(C T ), By Lemma 2.3 N(KA T K) N(KC T K) x 1 KC T K =0. Now, by using x 1 KA T K =0,x 1 KC T K = 0 and x 2 K(M/A) T K = 0 it can be verified that (x 1 x 2 ) KA T K KC T K 0 K(M/A) T K =0. Thus N(M 1 ) N(KM T 1 K). By using Lemma(2.3) N(KMT 1 K) N(M 1). Therefore N(M 1 )=N(KM T 1 K). Hence M 1 is k-kernel symmetric. In the same manner, it can be proved that M 2 is k-kernel symmetric. Thus (2) (3) holds. (3) (1): M 1 is k - Kernel symmetric N(M 1 )=N(KM T 1 K) M 2 is k - Kernel symmetric N(M 2 )=N(KM T 2 K)

6 336 A. R. Meenakshi and D. Jaya Shree To prove, M is k-kernel Symmetric that is N(M) =N(KM T K). Let x N(M) xm =0. Partition x in conformity with that of M as x = x 1 x 2 then, (x 1 x 2 ) =0. x 1 A + x 2 C =0 x 1 A = 0 and x 2 C =0 x 1 B + x 2 D =0 x 1 B = 0 and x 2 D =0 From the definition of M/A = D - CA B we have, x 2 D = 0 and x 2 C =0 x 2 (M/A) =0 x 1 A+x 2 C = 0 and x 2 (M/A) = 0 (3.4) And x 1 A =0,x 1 B+x 2 (M/A) = 0 (3.5) From (3.4), x N(M 1 ) x N(KM T 1 K). From (3.5), x N(M 2 ) x N(KM T 2 K).Hence x N(KMT K). N(M) N(KM T K).Similarly N(KM T K) N(M). Therefore M is k - Kernel symmetric matrix.hence the theorem. Theorem 3.6 Let M be a matrix of the form (3.1) with N(A T ) N(C T ) and N((M/A) T ) N(B T ), then the following are equivalent. (1) M is k - Kernel symmetric matrix with k = k 1 k 2. (2) A is k -Kernel symmetric, M/A is k -Kernel symmetric further N(A) N(B)and N(M/A) N(C) (3) Both the matrices and are k - Kernel symmet- C M/ M/A

7 On Schur complement 337 Proof ric. This theorem follows immediately from Theorem (3.2) and from the fact that M is k - Kernel symmetric M T is k - Kernel symmetric. Corollary 3.7 A C T Let M be a matrix of the form with N(A) N(C T ) and N((M/A)) N(C),then the following are equivalent. (1) M is k - Kernel symmetric matrix with k = k 1 k 2. (2) A is k -Kernel symmetric, M/A is k -Kernel symmetric (3) The matrix is k - Kernel symmetric. C M/A Remark 3.8 The condition taken on M in Theorem 3.2 and Theorem 3.6 are essential. This is illustrated in the following example. Example: Let M = and K = For this M, since M has ho zero rows and no zero columns N(M) = {0}. N(KM T K)={0}. Thus N(M) = N(KM T K) Mis k - Kernel Symmetric. A = M/A = D CA B = is a g-inverse, with respect to A M/A is k - Kernel Symmetric, since N(M/A) =N(K(M/A) T K)={0}. Aisk - Kernel Symmetric, since N(A) =N(KA T K)={0} for all K.

8 338 A. R. Meenakshi and D. Jaya Shree N(A) N(B) and N(A T ) N(C T ). Here, N(M/A) = {(0,x 2 ):x 2 F}= N((M/A) T ), N(C) = {0}, N(B T )={0}. N(M/A) not contained in N(C) and N((M/A) T ) not contained in N(B T ). Further, M 1 = x 4 F} N(M 1)={0}. N(KM1 T K)={(0, 0, 0,x 4): M 1 is not k- Kernel Symmetric. M 2 = M 2 is not k- Kernel Symmetric. Thus condition (1) of Theorem (3.2) holds but condition (2) and (3) of the Theorem (3.2) fail. Thus condition (1) of Theorem (3.6) holds but condition (2) and (3) fail. Remark 3.9 For a k - Kernel symmetric matrix M of the form M 1 = k 1 k 2, the following are equivalent N(A) N(B),N(M/A) N(C) N(A T ) N(C T ),N((M/A) T ) N(B T ) with k = However this fails if we omit the condition that M is k -Kernel symmetric. Example:

9 On Schur complement 339 M = and K = N(M) ={(0, 0,x 3, 0) : x 3 F} N(KM T K). Therefore M is not k - Kernel symmetric. 0 0 M/A =. Here N(A) N(B),N(M/A) N(C) but N(A T ) is not 1 0 contained in N(C T ),N((M/A) T )is not contained in N(B T ). References 1 Carlson D, Haynsworth E and Markham TH, A generalization of the schur complement by means of the Moore Penrose inverse, SIAM J.APPL.Math 26 (1974) Hill R.D and Waters S.R, On k-real and k-hermitian matrices, Lin.Alg.and its Appln.,169 (1992) Kim,K.H and Roush,F.W., On Generalized fuzzy matrices,fuzzy Sets and Systems, 1980,4: AR.Meenakshi, Fuzzy matrix:theory and Applications,MJP Publishers,Chennai Meenakshi AR and Jaya Shree D, On k-kernel Symmetric matrices, volume 2009 (2009), Article ID , 8 pages. 6 Meenakshi AR and Krishnamoorthy S, On k-ep matrices,lin.alg.and its Appln.,269 (1998) Meenakshi AR and Krishnamoorthy S, On Schur complement in k-ep matrices, Indian J.Pure appl.math (2002) Received: June, 2009